Question

measurements show that the energy of a mixture of gaseous reactants decreases by 116. kj during a certain chemical reaction, which is carried out at a constant pressure. furthermore, by carefully monitoring the volume change it is determined that - 74. kj of work is done on the mixture during the reaction. calculate the change in enthalpy of the gas mixture during the reaction. be sure your answer has the correct number of significant digits. is the reaction exothermic or endothermic? exothermic endothermic

Explanation:

Step1: Recall the first law of thermodynamics and enthalpy change at constant pressure

The first law of thermodynamics is \(\Delta U = q + w\), and at constant pressure, the enthalpy change \(\Delta H=\Delta U\) when only pressure - volume work is considered? Wait, no, actually, at constant pressure, \(\Delta H = q_p\) (heat at constant pressure). And from the first law, \(\Delta U=q + w\). But we know that the energy of the reactants (the internal energy of the system) decreases by 116 kJ, so \(\Delta U=- 116\space kJ\) (since the system's energy is decreasing). The work done on the mixture is \(w = - 74\space kJ\)? Wait, no: the sign convention: if work is done on the system, \(w\) is positive; if work is done by the system, \(w\) is negative. Wait, the problem says " - 74 kJ of work is done on the mixture", so \(w=-74\space kJ\)? Wait, no, let's clarify the sign convention. Let's use the convention where:

- \(\Delta U\): change in internal energy of the system. If the system's energy decreases, \(\Delta U\) is negative.
- \(q\): heat added to the system (positive if heat is added, negative if heat is lost).
- \(w\): work done on the system (positive if work is done on the system, negative if work is done by the system).

The problem states that the energy of the gaseous reactants (system) decreases by 116 kJ, so \(\Delta U=- 116\space kJ\). The work done on the mixture (system) is - 74 kJ? Wait, no, the problem says " - 74 kJ of work is done on the mixture", so according to the sign convention, if work is done on the system, \(w\) should be positive. Wait, maybe the problem has a different sign convention. Let's re - read: " - 74 kJ of work is done on the mixture". So if work is done on the system, \(w = + 74\space kJ\)? Wait, no, the negative sign in " - 74 kJ" might be indicating the direction. Wait, maybe the standard sign convention here is: \(w\) is positive when work is done on the system, negative when done by the system. So if the work done on the mixture is - 74 kJ, that means the mixture (system) is doing work, \(w=-74\space kJ\).

But we know that at constant pressure, \(\Delta H = q_p\), and from the first law \(\Delta U=q + w\). We know \(\Delta U=-116\space kJ\) (energy of reactants decreases, so \(\Delta U\) of the system is negative) and \(w=-74\space kJ\) (work done on the system is - 74 kJ, i.e., work done by the system is 74 kJ). Then we can solve for \(q\):

\(q=\Delta U - w\)

Substitute \(\Delta U=-116\space kJ\) and \(w = - 74\space kJ\) into the formula:

\(q=-116-( - 74)=-116 + 74=-42\space kJ\)? Wait, that can't be right. Wait, maybe I mixed up the sign of \(\Delta U\). Wait, the energy of the reactants (system) decreases by 116 kJ, so the internal energy of the system (\(\Delta U\)) is equal to the final energy minus the initial energy. If the initial energy is \(U_i\) and final is \(U_f\), \(\Delta U=U_f - U_i\). If the energy decreases, \(U_f<U_i\), so \(\Delta U=- 116\space kJ\) (correct).

Wait, another approach: Enthalpy change at constant pressure \(\Delta H=\Delta U + P\Delta V\). And the work done in a pressure - volume process is \(w=-P\Delta V\) (when work is done by the system, \(w\) is negative; when work is done on the system, \(w\) is positive). Wait, the problem says " - 74 kJ of work is done on the mixture", so \(w = - 74\space kJ\) (using the convention \(w=-P\Delta V\), so if work is done on the system, \(P\Delta V\) is negative, so \(w=-P\Delta V\) is positive? No, let's use the IUPAC sign convention:

- \(q\): positive if heat is absorbed by the system (endothermic), negative if heat is released (exothermic).
- \(w\): positive if work is done on the system, negative if work is done by the system.

The internal energy change \(\Delta U=q + w\).

The energy of the gaseous reactants (system) decreases by 116 kJ, so \(\Delta U=-116\space kJ\) (because the system's energy is decreasing).

The work done on the mixture (system) is - 74 kJ, which means work is done by the system, so \(w=-74\space kJ\) (since work done by the system is negative in this convention).

Now, we want to find \(\Delta H\) at constant pressure. At constant pressure, \(\Delta H = q_p\), and from \(\Delta U=q + w\), we can solve for \(q\):

\(q=\Delta U - w\)

Substitute \(\Delta U=-116\space kJ\) and \(w=-74\space kJ\):

\(q=-116-(-74)=-116 + 74=-42\space kJ\)? Wait, that's not matching. Wait, maybe the energy decrease of the reactants is the \(\Delta U\) of the system. Wait, no, the reactants are the system, so when the reaction occurs, the internal energy of the system (reactants) decreases by 116 kJ, so \(\Delta U=-116\space kJ\). The work done on the system is - 74 kJ (so work done by the system is 74 kJ, \(w = - 74\space kJ\)). Then \(q=\Delta U - w=-116-(-74)=-42\space kJ\). But at constant pressure, \(\Delta H = q_p\), so \(\Delta H=-42\space kJ\)? Wait, that seems wrong. Wait, maybe the sign of \(w\) is different. Let's assume that the work done on the system is positive, so \(w = + 74\space kJ\) (because the problem says " - 74 kJ of work is done on the mixture", maybe the negative sign is a mistake, or maybe the sign convention is reversed). Let's try that:

If \(w = + 74\space kJ\) (work done on the system) and \(\Delta U=-116\space kJ\) (system's energy decreases), then from \(\Delta U=q + w\), we have \(q=\Delta U - w=-116 - 74=-190\space kJ\), which also doesn't make sense.

Wait, maybe we should use the definition of enthalpy: \(H = U+PV\), so \(\Delta H=\Delta U+\Delta(PV)\). At constant pressure, \(\Delta(PV)=P\Delta V\), so \(\Delta H=\Delta U + P\Delta V\). And the work done in a pressure - volume process is \(w=-P\Delta V\) (work done by the system is \(P\Delta V\), so \(w=-P\Delta V\) when work is done by the system, \(w = P\Delta V\) when work is done on the system).

The problem states that the energy of the gaseous reactants (system) decreases by 116 kJ, so \(\Delta U=-116\space kJ\). The work done on the mixture is - 74 kJ, so \(w=-74\space kJ\) (which means work is done by the system, so \(w=-P\Delta V=-74\space kJ\), so \(P\Delta V = 74\space kJ\)).

Then \(\Delta H=\Delta U+P\Delta V=-116 + 74=-42\space kJ\)? No, that's not right. Wait, maybe the energy decrease of the reactants is the \(\Delta H\)? Wait, no, the enthalpy change of the reaction is related to the heat at constant pressure.

Wait, let's start over. The key formula: At constant pressure, \(\Delta H = q_p\), and from the first law of thermodynamics, \(\Delta U=q + w\).

We know:

- The energy of the gaseous reactants (system) decreases by 116 kJ, so \(\Delta U=-116\space kJ\) (because the system's internal energy is decreasing).
- Work done on the mixture (system) is - 74 kJ. So, using the sign convention where work done on the system is positive, work done by the system is negative. So if work done on the system is - 74 kJ, that means work is done by the system, so \(w=-74\space kJ\).

Now, solve for \(q\) from \(\Delta U = q+w\):

\(q=\Delta U - w\)

Substitute \(\Delta U=-116\space kJ\) and \(w = - 74\space kJ\):

\(q=-116-(-74)=-116 + 74=-42\space kJ\)

But at constant pressure, \(\Delta H = q_p\), so \(\Delta H=-42\space kJ\)? Wait, but the problem says the energy of the reactants decreases by 116 kJ. Maybe I messed up the sign of \(\Delta U\). If the reactants are the system, and after the reaction, the products have lower energy? Wait, no, the reactants' energy decreases by 116 kJ, so the change in internal energy of the system (reactants) is \(\Delta U=-116\space kJ\).

Wait, maybe the correct approach is: The enthalpy change of the system (gas mixture) is equal to the heat exchanged at constant pressure. And we know that \(\Delta U = q + w\). We are given that the energy of the reactants (system) decreases by 116 kJ, so \(\Delta U=-116\space kJ\). The work done on the system is - 74 kJ (so \(w=-74\space kJ\)). Then \(q=\Delta U - w=-116-(-74)=-42\space kJ\). But since at constant pressure, \(\Delta H = q\), so \(\Delta H=-42\space kJ\)? No, that can't be. Wait, maybe the work done on the system is + 74 kJ. Let's try that. If work done on the system is \(w = + 74\space kJ\) (the - 74 kJ in the problem is a mistake, or the sign convention is different). Then:

\(q=\Delta U - w=-116 - 74=-190\space kJ\), which is also not right.

Wait, maybe the energy decrease of the reactants is the \(\Delta H\). Wait, the problem says "the energy of a mixture of gaseous reactants decreases by 116 kJ". Enthalpy change of the reaction (for the system) is related to the heat at constant pressure. If the reactants' energy decreases, and we consider the reaction, maybe \(\Delta H=-116\space kJ\)? But then what about the work?

Wait, the formula for enthalpy is \(H = U+PV\), so \(\Delta H=\Delta U+\Delta(PV)\). At constant pressure, \(\Delta(PV)=P\Delta V\), and the work done \(w=-P\Delta V\) (work done by the system). So \(\Delta H=\Delta U - w\) (because \(w=-P\Delta V\), so \(P\Delta V=-w\), then \(\Delta H=\Delta U - w\)).

We know \(\Delta U=-116\space kJ\) (energy of reactants decreases), and \(w=-74\space kJ\) (work done on the system is - 74 kJ, so work done by the system is 74 kJ, \(w=-P\Delta V=-74\space kJ\), so \(P\Delta V = 74\space kJ\)). Then \(\Delta H=\Delta U - w=-116-(-74)=-116 + 74=-42\space kJ\). But this seems incorrect. Wait, maybe the question is simpler: at constant pressure, \(\Delta H = q_p\), and the change in internal energy \(\Delta U = q+w\). But if the energy of the reactants (system) decreases by 116 kJ, that is \(\Delta U=-116\space kJ\). The work done on the system is \(w = - 74\space kJ\) (work done by the system is 74 kJ). Then \(q=\Delta U - w=-116-(-74)=-42\space kJ\). So \(\Delta H = q=-42\space kJ\)? But the number of significant figures: 116 has 3, 74 has 2, so the answer should have 2 significant figures? Wait, 116 is three, 74 is two. When subtracting or adding, the number of decimal places matters, but here we have whole numbers. Wait, 116 - 74 = 42, but with signs. Wait, maybe the correct answer is - 42 kJ? But let's check the exothermic/endothermic part. If \(\Delta H\) is negative, the reaction is exothermic (releases heat).

Wait, maybe I made a mistake in the sign of \(w\). Let's use the following sign convention:

- \(q\): positive when heat is absorbed by the system (endothermic), negative when heat is released (exothermic).
- \(w\): positive when work is done on the system, negative when work is done by the system.

The internal energy change \(\Delta U=q + w\).

The system (gaseous reactants) has a decrease in energy, so \(\Delta U=-116\space kJ\) (system loses energy).

Work done on the system: \(w=-74\space kJ\) (so work is done by the system, \(w\) is negative).

Then:

\(q=\Delta U - w=-116-(-74)=-116 + 74=-42\space kJ\)

Since \(q\) is negative, heat is released by the system, so the reaction is exothermic. And at constant pressure, \(\Delta H = q=-42\space kJ\)? But this seems to ignore the work? Wait, no, at constant pressure, \(\Delta H = q_p\), and \(q_p=\Delta U - w\) (because \(w=-P\Delta V\) and \(\Delta H=\Delta U+P\Delta V=\Delta U - w\)).

Wait, maybe the problem is that the "energy of the gaseous reactants" is the internal energy, and we are to find the enthalpy change. So \(\Delta H=\Delta U+P\Delta V\), and \(w=-P\Delta V\), so \(\Delta H=\Delta U - w\). We have \(\Delta U=-116\space kJ\) and \(w=-74\space kJ\), so \(\Delta H=-116-(-74)=-42\space kJ\).

For the exothermic/endothermic part: since \(\Delta H\) is negative (or \(q\) is negative, heat is released), the reaction is exothermic.

Step 2: Calculate \(\Delta H\)

We know that \(\Delta U=-116\space kJ\) (energy of reactants decreases), \(w = - 74\space kJ\) (work done on the system is - 74 kJ, i.e., work done by the system is 74 kJ). Using the formula \(\Delta H=\Delta U - w\) (at constant pressure, \(\Delta H = q_p\) and from \(\Delta U=q + w\) and \(w=-P\Delta V\), \(\Delta H=\Delta U+P\Delta V=\Delta U - w\)):

\(\Delta H=-116-(-74)=-116 + 74=-42\space kJ\)

Step 3: Determine if the reaction is exothermic or endothermic

Since \(\Delta H\) (or \(q_p\)) is negative, heat is released by the system to the surroundings. A reaction that releases heat is exothermic.

Answer:

The change in enthalpy is \(\boldsymbol{-42}\) kJ, and the reaction is exothermic.